In the following post by DavidLHarden : See Here

He quoted the following claim: "There is a theorem that says that if $p$ is a prime and $|G|=p^n $ , then $|AutG| $ divides $ \Pi_{k=0}^{n-1} (p^{n}-p^{k}) $ " .

I can't find any reference for this theorem ,

Does someone know of any reference for this fact?

Thanks in advance

  • $\begingroup$ I think the theorem in this form may be due to Peter M. Neumann- so try Mathscinet if you have access- assuming the theorem is stated correctly. It is certainly true that ${\rm Aut}(G)$ has a normal $p$-subgroup whose index is bounded by the number you quote (this is a Theorem of W.Burnside). $\endgroup$ – Geoff Robinson Nov 23 '12 at 8:28
  • $\begingroup$ I don't have Khukhro's book "$p$-Automorphisms of Finite $p$-Groups" here, but it would be a good place to try. $\endgroup$ – M T Nov 23 '12 at 10:38
  • $\begingroup$ Geoff: I would try Mathscinet on Monday or something... I am not sure if this is due to Neumann or not... Thanks anyway! @mt: I couldn't find this theorem in the book you just mentioned... Thanks for the suggestion! $\endgroup$ – Jason Mraz Nov 23 '12 at 19:16

I don't have a reference, but here's the next best thing: a proof.

First, let's fix some notation. Let $P$ be a p-group, $G$ it's group of automorphisms, and $\Phi(P) = P^p[P,P]$ it's Frattini subgroup. Define inductively $\Phi^k(P)$ as $\Phi(\Phi^{k-1}(P)).$

The subgroups $\Phi^k(P)$ form a decreasing chain of characteristic subgroups which exhaust $P.$

Let $$G_k := \ker(G \rightarrow Aut(P/\Phi^k(P)))$$ and $$G_k' := \ker(G \rightarrow Aut(\Phi^k(P)/\Phi^{k+1}(P))).$$

Then the subgroups $G_k$ form an decreasing chain of normal subgroups of $G$ which exhaust $G.$

Let $d_k = \dim_{\mathbb{F}_p}(\Phi^k(P)/\Phi^{k+1}(P)).$ The group $P$ can be generated by $d_0$ elements. Choose a generating set $g_1 ... g_{d_0}$ and consider the map from $G_k \cap G_k'/G_{k+1}$ to $(\Phi^k(P)/\Phi^{k+1}(P))^{d_0}$

given by

$$\sigma \mapsto (\sigma(g_i)g_i^{-1})_{i=1}^{d_0}.$$

This map is an injective group homomorphism.

On the other hand $G_k/(G_{k} \cap G_{k}')$ injects into $Aut(\Phi^k(P)/\Phi^{k+1}(P)) \cong GL_{d_i}( \mathbb{F}_p).$

Note that if $p^n = |P|$ and $r = max\{d_i : d_i \neq 0\},$ then $\displaystyle\sum_{i=0}^r d_i = n.$ It follows that the order $$|G| = \displaystyle\prod_{k=0}^{r}|G_k/G_{k+1}| = \displaystyle\prod_{k=0}^{r} |(G_k \cap G_k')/G_{k+1}||(G_k \cap G_k')/G_{k+1}|$$ divides

$$\displaystyle\prod_{s=0}^{d_0-1} (p^{d_0} - p^s)\displaystyle\prod_{k=1}^{r} p^{d_kd_0}\displaystyle\prod_{s=0}^{d_k-1} (p^{d_k} - p^s)$$

which divides

$$\displaystyle\prod_{k=0}^{n-1} (p^n - p^k).$$

  • 1
    $\begingroup$ @JSpecter. Since $\prod(p^n-p^k)$ is the order of GL$(n,p)$, I'm wondering if Aut$(G)$ is in fact isomorphic to some subgroup of GL$(n,p)$. Is it a silly thought? $\endgroup$ – Brendan McKay Nov 24 '12 at 0:32
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    $\begingroup$ @McKay, when $p>2$, the automorphism group of a cyclic group of order $p^3$ contains elements of order $p^2$, but $GL(3,p)$ does not. $\endgroup$ – Rene Schoof Nov 24 '12 at 14:56
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    $\begingroup$ Why is a reference considered preferable to a proof? A proof is what you want to find in the reference. $\endgroup$ – DavidLHarden Nov 25 '12 at 11:15
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    $\begingroup$ @DavidLHarden: But you don't want to prove something that is well known in a paper you are writing... By proving it, it's kind of saying that you are the one that figured this theorem out... But if you already saw this post... Have you got any idea for possible reference? In your first message I quoted, you said that it's well known, do you know where can I find it ? Thanks ! $\endgroup$ – Jason Mraz Nov 25 '12 at 22:02
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    $\begingroup$ There is some material that is pretty close to a proof of this (it contains most, if not all, of the elements of the proof) in Marshall Hall Jr.'s "Theory of Finite Groups". $\endgroup$ – DavidLHarden Nov 26 '12 at 2:04

As @DavidLHarden explains in the link that you gave, this theorem is proved by attending to the $p$-part and $p'$-part separately.

For the $p'$-part the result follows from the following theorem of Burnside:

Let $\psi$ be a $p'$-automorphism of the $p$-group $P$ which induces the identity on $P/\Phi(P)$. Then $\psi$ is the identity automorphism of $P$.

This is the result that Geoff refers to in his comment above. It is discussed and proved in Section 5 of Gorenstein's Finite Groups, specifically Theorem 1.4 of that section.

I do not know of a reference for the $p$-part of the proof. You should certainly look at the paper by Neumann that Geoff mentions, however if I understand that proof correctly it only proves your bound for $|Out P|$, rather than $|Aut P|$. On the other hand Neumann is considering a much more general setting than just $p$-groups.

  • $\begingroup$ @Nick Gill: Thanks ! Sorry for my ignorance, but what is the $p'$ part of the result? (I can't understand what is $p'$ in your response) . I also have trouble getting into the link you just gave... It gives me an error... can you please fix it? Thanks ! $\endgroup$ – Jason Mraz Nov 23 '12 at 19:26
  • $\begingroup$ @Jason, I've fixed the link. Sorry about that. You can split this bound into two parts: prove that the biggest power of p dividing Aut G divides the given product, and then prove that the biggest integer coprime to p dividing Aut G divides the given product. When I write the p'-part I mean the latter. $\endgroup$ – Nick Gill Nov 23 '12 at 21:31

In fact, if $|G|=p^n$ and $d(G)=d$, then $|\mathrm{Aut}(G)|$ divides the number $$(p^n-p^{n-d})(p^n-p^{n-d+1}) ... (p^n-p^{n-1}).$$ This result is due to P. Hall (1933)

Proof. The above product is the number of minimal bases of $G$. However, that number is a multiple of $|\text{Aut}(G)|$. Indeed, if $\cal B$ is the set of all bases of $G$, then all $\text{Aut}(G)$-orbits on $\cal B$ have the same cardinality $|\text{Aut}(G)|$.

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    $\begingroup$ Could you add a more precise reference, please? $\endgroup$ – Sebastian Goette Dec 18 '15 at 21:36

It's actually easy to put a bound on $Aut(P)$ for some finite $p$-group $P$, simply using Burnside's basis theorem, which says every basis for the elementary abelian group $P/\Phi(P)$ corresponds to a minimal generating set for the group $P$. Clearly, any element of $Aut(P)$ must take one minimal generating set to another.

So how many minimal generating sets are there? Well, if $|\Phi(P)| = p^d$, and $|P/\Phi(P)|=p^e$, then a minimal generating set consists of $e$ elements. There are $$ \prod_{k=0}^{e-1} (p^e-p^k) $$ different bases of $P/\Phi(P)$ (as a vector space of dimension $e$). Each such element really represents a coset of $\Phi(P)$, which contains $p^d$ elements; that is, for such a given basis, each basis vector has $p^d$ choices up in $P$; all told, then, there are $$ p^{de} \prod_{k=0}^{e-1} (p^e-p^k) $$

minimal generating sets. Of course, $Aut(P)$ acts on these freely (as it is defined by what it does to a generating set), so its order divides that number.


In post 6 is presented a weak result (it attained only for elementary abelian $P$). The Hall's result in post 3 is attained by many groups. It may be improved only in the case if we have an additional information on $P$.


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